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How do you work with the three trigonometric functions, the sine and cosine rules, and exact values to model and solve triangle problems?

The sine, cosine and tangent functions and their graphs, the sine and cosine rules, the area of a triangle, and exact values of trigonometric ratios for standard angles.

A focused answer to the OCR A-Level Mathematics A trigonometry content, covering the sine, cosine and tangent functions and their graphs, the sine and cosine rules, the area of a triangle, the ambiguous case, and the exact values of trigonometric ratios for standard angles.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

OCR wants you to understand the sine, cosine and tangent functions and their graphs (including period, symmetry and key values), use the sine and cosine rules to solve any triangle, find the area of a triangle from two sides and the included angle, deal with the ambiguous case of the sine rule, and recall the exact values of the trigonometric ratios for the standard angles.

The answer

The three functions and their graphs

The functions sinθ\sin\theta and cosθ\cos\theta are periodic with period 360360^\circ (or 2π2\pi radians), oscillating between 1-1 and 11. The function tanθ\tan\theta has period 180180^\circ (or π\pi) and vertical asymptotes where cosθ=0\cos\theta = 0, that is at θ=90,270,\theta = 90^\circ, 270^\circ, \dots Knowing the shape of each graph lets you read off how many solutions an equation has in a given interval and find all of them using the symmetry of the curve.

Exact values

You must know the exact ratios for 00^\circ, 3030^\circ, 4545^\circ, 6060^\circ and 9090^\circ without a calculator. They come from the half-equilateral triangle and the unit square.

The sine rule

The sine rule links each side of a triangle with the angle opposite it. Use it when you know two angles and any side, or two sides and a non-included angle.

The cosine rule

The cosine rule is the right tool when you know two sides and the included angle (to find the third side), or all three sides (to find an angle).

Area of a triangle

When you know two sides and the angle between them, the area is 12absinC\tfrac{1}{2}ab\sin C, where CC is the included angle. This is faster and more accurate than finding a perpendicular height.

Examples in context

Choosing the right rule

The decision is mechanical. If you have an angle and the side opposite it, the sine rule works. If you have two sides and the angle between them, or three sides, use the cosine rule. Many multi-step questions need the cosine rule first to find a side, then the sine rule or the area formula to finish.

The ambiguous case

When the sine rule gives an angle from a side opposite a known acute angle, there can be two triangles: the calculator returns the acute value, but the obtuse supplement 180θ180^\circ - \theta may also fit. Always check whether the obtuse option keeps the angle sum below 180180^\circ; if it does, both triangles are valid and the question usually asks for both.

Try this

Q1. A triangle has sides 44 cm and 55 cm with an included angle of 6060^\circ. Find its area. [2 marks]

  • Cue. 12(4)(5)sin60=10×32=538.66\tfrac{1}{2}(4)(5)\sin 60^\circ = 10 \times \tfrac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66 cm2^2.

Q2. In triangle ABCABC, a=8a = 8, b=11b = 11 and c=7c = 7. Find angle AA. [3 marks]

  • Cue. cosA=112+72822(11)(7)=106154=0.688\cos A = \dfrac{11^2 + 7^2 - 8^2}{2(11)(7)} = \dfrac{106}{154} = 0.688, so A46.5A \approx 46.5^\circ.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20195 marksIn triangle ABCABC, AB=7AB = 7 cm, AC=9AC = 9 cm and angle BAC=50BAC = 50^\circ. Find the length BCBC and the area of the triangle.
Show worked answer →

Use the cosine rule to find BCBC (M1): BC2=72+922(7)(9)cos50BC^2 = 7^2 + 9^2 - 2(7)(9)\cos 50^\circ.

BC2=49+81126cos50=130126(0.6428)=13080.99=49.01BC^2 = 49 + 81 - 126\cos 50^\circ = 130 - 126(0.6428) = 130 - 80.99 = 49.01 (A1), so BC=49.017.00BC = \sqrt{49.01} \approx 7.00 cm (A1).

Area =12absinC=12(7)(9)sin50= \tfrac{1}{2}ab\sin C = \tfrac{1}{2}(7)(9)\sin 50^\circ (M1) =31.5×0.76624.1= 31.5 \times 0.766 \approx 24.1 cm2^2 (A1).

Markers reward selecting the cosine rule for the included-angle case, evaluating BCBC, the area formula with the included angle, and the final area.

OCR 20214 marksIn triangle PQRPQR, PQ=10PQ = 10 cm, angle QPR=40QPR = 40^\circ and angle PQR=75PQR = 75^\circ. Find the length PRPR.
Show worked answer →

The third angle is PRQ=1804075=65PRQ = 180^\circ - 40^\circ - 75^\circ = 65^\circ (M1).

Apply the sine rule with the side PRPR opposite angle QQ (M1): PRsin75=10sin65\dfrac{PR}{\sin 75^\circ} = \dfrac{10}{\sin 65^\circ} (A1).

So PR=10sin75sin65=10(0.9659)0.906310.7PR = \dfrac{10\sin 75^\circ}{\sin 65^\circ} = \dfrac{10(0.9659)}{0.9063} \approx 10.7 cm (A1).

Markers reward finding the third angle, pairing each side with its opposite angle in the sine rule, and the correct length.

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